Prove that $5^{2^n} – 1$ is divisible by $2^{n+1}$ for all $n ≥ 1$

divisibilityelementary-number-theoryinduction

I made the following using induction:

If $n=1$ then the proposition is true: $5^{2^1} – 1=24$ is divisible by $2^{1+1} = 4$

Now I suppose that for a natural number $k$, $5^{2^k} – 1$ is divisible by $2^{k+1}$ is true. And I want to prove (using this) that the proposition is true for $n=k+1$ but I don't know how to do this.

I appreciate the help you give me.

Best Answer

We have:

$$5^{2^{k+1}} - 1 = \left(5^{2^k} - 1\right)\left(5^{2^k}+1\right)$$

By induction the first factor is divisible by $2^{k+1}$, while the second one is obviously even and hence $5^{2^{k+1}} - 1$ is divisible by $2^{k+2}$

Related Question